(1/3)^x = 81?

Answer 1

(1/3)^x = 3^(-x)

Does that help?

If not -- then 9*9 =81, so 3*3*3*3 = 3^4 = 81

This gives us -x = 4

x = -4

Answer 2

The question basically asks: To what power can you raise (1/3) so that you get 81?

On the one hand, If you start thinking: 1,2,3,etc, you get into a problem... (1/3)^1 = (1/3); (1/3)^2=1/9, (1/3)^3=1/27... You see, your answer is getting smaller and smaller, not bigger and bigger, which means you will not get to 81 that way.

On the middle hand, (1/3)^0 = 1.

And on the other hand, whenever a fraction is raised to a negative power, the numerator and the denominator of the fraction change places. That is, (1/3)^(-1) = (3/1)^1=3^1=3. Similarly, (1/3)^(-2)=(3/1)^2=3^2=9. This gives you the clue that the exponent is negative: so that the sequence keeps growing and growing... Knowing that, all you really need to know is this: to what power can I raise 3 so that I get 81? I assume that it should be easy potatoes to come up with 4: 3*3*3*3=81.

So, therefore, the answer is x=-4.

Yes, I am aware that you handicapped humans only have 2 hands.

Answer 3

(1/3)^x = 81
log(1/3)81 = x

x = (log(81))/(log(1/3))
x = -4

ANS : -4

Another way of putting this is

(1/3)^x = 81
3^(-x) = 81
3^(-x) = 3^4
-x = 4
x = -4

Answer 4

(1/3)^x = 81
3^(-x) = 81
3^(-x) = 3^4
-x = 4
x = -4

Answer 5

dammit, do your own homework... jk...

3 ^ 4 = 81

1 / 3 = 3 ^ -1

(3 ^ - 1)^x = 3 ^ 4
3 ^-x = 3^4

x = -4

Answer 6

x = negative 4

Answer 7

(1/3)^x=1/3^-4
x=-4

Answer 8

3^(-x)=81
3^(-x)=3^4
-x=4
x=-4